Physics - Capacitor Question with Solution | TestHub

When switch is at position-1

$q_{\max }=C{\epsilon }_1$

$U_i=\frac{1}{2}C{\epsilon }_1^2$

$\frac{\left(C{\epsilon }_1-q\right)}{C}={\epsilon }_2-\frac{dq}{dt}R=0$

$({\epsilon }_1-{\epsilon }_2)-\frac{q}{C}=\frac{dq}{dt}\times R$

$C({\epsilon }_1+{\epsilon }_2)-q=RC\frac{dq}{dt}$

${\int }_0^q\frac{dq}{C({\epsilon }_1+{\epsilon }_2)-q}={\int }_0^t\frac{dt}{RC}$

$\ln \left[\frac{C({\epsilon }_1+{\epsilon }_2)-q}{C({\epsilon }_1+{\epsilon }_2)}\right]=-\frac{t}{RC}$

$1-\frac{q}{C({\epsilon }_1+{\epsilon }_2)}=e^{-\frac{t}{RC}}$

$\Rightarrow \left(1-e^{-\frac{t}{RC}}\right)\times C({\epsilon }_1+{\epsilon }_2)=q$

$i=\frac{dq}{dt}=\frac{C({\epsilon }_1+{\epsilon }_2)}{RC}{e}^{-\frac{t}{RC}}$

$i=\frac{({\epsilon }_1+{\epsilon }_2)}{R}{e}^{-\frac{t}{RC}}$

$\Delta H=\int \frac{({\epsilon }_1+{\epsilon }_2{)}^2}{R^2}\times R{e}^{-\frac{2t}{RC}}dt$

$=\frac{C({\epsilon }_1+{\epsilon }_2{)}^2}{2}[1-0]=\frac{C({\epsilon }_1+{\epsilon }_2{)}^2}{2}$